*3.3. The Block Tridiagonal Matrix*

The matrix form of a linearized tridiagonal system is:

$$\mathbf{A}\boldsymbol{\delta} = \mathbf{r},\tag{41}$$

where

$$\mathbf{S} = \begin{bmatrix} \begin{bmatrix} A\_1 \end{bmatrix} & \begin{bmatrix} \mathbf{C}\_1 \end{bmatrix} \\\\ \begin{bmatrix} B\_2 \end{bmatrix} & \begin{bmatrix} A\_2 \end{bmatrix} & \begin{bmatrix} \mathbf{C}\_2 \end{bmatrix} \end{bmatrix} \\\\ \mathbf{S} = \begin{bmatrix} & & & & & & & \\ & & & & & & \\ & & & & & & \\ & & & & & & \\ & & & & & & \\ & & & & & & \\ & & & & & \begin{bmatrix} A\_{f-1} \end{bmatrix} & \begin{bmatrix} A\_{f-1} \end{bmatrix} & \begin{bmatrix} \begin{bmatrix} & & & & \\ & \begin{bmatrix} & \\ \end{bmatrix} & \\\ \begin{bmatrix} & & & \\ \end{bmatrix} & \begin{bmatrix} & & & & \\ \end{bmatrix} \end{bmatrix}, \mathbf{r} = \begin{bmatrix} & & & & & \\ & \begin{bmatrix} & & & \\ \end{bmatrix} & & & \\ & & & & \begin{bmatrix} & & & \\ \end{bmatrix} & \begin{bmatrix} & & & & \\ \end{bmatrix} \end{bmatrix}, \mathbf{r} = \begin{bmatrix} & & & & \\ & \begin{bmatrix} & & & \\ \end{bmatrix} & & & \\ & & & & \\ \end{bmatrix} \end{bmatrix}$$

The boundary conditions in Equation (32) are satisfied precisely with no iteration. Due to these suitable values being maintained in every iterate, we assume δ*F*0 = 0, δ*<sup>w</sup>*0 = 0, δ*p*0 = 0, <sup>δ</sup>*wJ* = 0, <sup>δ</sup>*gJ* = 0, and let *dJ* = −12 *hJ*.

The entries of the matrices are

$$\begin{aligned} \begin{bmatrix} A\_1 \end{bmatrix} &= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 \\ d\_1 & 0 & 0 & d\_1 & 0 \\ 0 & -1 & 0 & 0 & d\_1 \\ (a\_2)\_1 & (a\_8)\_1 & (a\_3)\_1 & (a\_1)\_1 & 0 \\ 0 & (b\_8)\_1 & (b\_3)\_1 & 0 & (b\_1)\_1 \end{bmatrix} \\\\ \begin{bmatrix} A\_j \\ -1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ (a\_6)\_j & (a\_8)\_j & (a\_3)\_j & (a\_1)\_j & 0 \\ (b\_6)\_j & (b\_8)\_j & (b\_3)\_j & 0 & (b\_1)\_j \end{bmatrix}, 2 \le j \le J, \end{aligned} \tag{43}$$

$$
\begin{bmatrix} B\_j \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & d\_j & 0 \\ 0 & 0 & 0 & 0 & d\_j \\ 0 & 0 & (a\_4)\_j & (a\_2)\_j & 0 \\ 0 & 0 & (b\_4)\_j & 0 & (b\_2)\_j \\ \end{bmatrix}, \\ 2 \le j \le l,\tag{44}
$$

$$
\begin{bmatrix} \mathbb{C}\_{j} \end{bmatrix} = \begin{bmatrix} d\_{j} & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \begin{pmatrix} a\_{5} \end{pmatrix}\_{j} & \begin{pmatrix} a\tau \end{pmatrix}\_{j} & 0 & 0 & 0 \\ \begin{pmatrix} a\tau \end{pmatrix}\_{j} & 0 & 0 & 0 \\ \begin{pmatrix} b\tau \end{pmatrix}\_{j} & 0 & 0 & 0 \end{pmatrix}, \tag{45}
$$

$$\begin{aligned} \left[\delta\_{1}\right] = \begin{bmatrix} \delta z\_{0} \\ \delta g\_{0} \\ \delta F\_{1} \\ \delta \boldsymbol{r}\_{1} \\ \delta p\_{1} \end{bmatrix} \cdot \left[\delta\_{j}\right] = \begin{bmatrix} \delta w\_{j-1} \\ \delta g\_{j-1} \\ \delta F\_{j-1} \\ \delta \boldsymbol{r}\_{j-1} \\ \delta p\_{j-1} \end{bmatrix} 2 \le j \le \mathsf{J}, \left[\mathsf{r}\_{j}\right] = \begin{bmatrix} (\mathsf{r}\_{1})\_{j-(1/2)} \\ (\mathsf{r}\_{2})\_{j-(1/2)} \\ (\mathsf{r}\_{3})\_{j-(1/2)} \\ (\mathsf{r}\_{4})\_{j-(1/2)} \\ (\mathsf{r}\_{5})\_{j-(1/2)} \end{bmatrix}, 1 \le j \le \mathsf{J} \end{aligned} \tag{46}$$

The final step is to solve the system in Equation (41) by the LU (lower–upper) factorization method, then implement numerical operations using MATLAB software (version 7, MathWorks, Natick, MA, USA). In this work the wall shear stress parameter *<sup>z</sup>*(*<sup>x</sup>*, 0) is considered as the convergence criterion (as it is usually considered, see Cebeci and Bradshaw [50]), so the calculations were repeated until the convergence criterion was satisfied, and stopped when <sup>δ</sup>*z*(*i*) 0 < ε1, where ε1 is chosen to be 10−<sup>5</sup> which give precise values up to four decimal places.
